Problem: A few families took a trip to an amusement park together. Tickets cost $$7.50$ each for adults and $$3.00$ each for kids, and the group paid $$45.00$ in total. There were $8$ fewer adults than kids in the group. Find the number of adults and kids on the trip.
Let $x$ equal the number of adults and $y$ equal the number of kids. The system of equations is then: ${7.5x+3y = 45}$ ${x = y-8}$ Solve for $x$ and $y$ using substitution. Since $x$ has already been solved for, substitute ${y-8}$ for $x$ in the first equation. ${7.5}{(y-8)}{+ 3y = 45}$ Simplify and solve for $y$ $ 7.5y-60 + 3y = 45 $ $ 10.5y-60 = 45 $ $ 10.5y = 105 $ $ y = \dfrac{105}{10.5} $ ${y = 10}$ Now that you know ${y = 10}$ , plug it back into ${x = y-8}$ to find $x$ ${x = }{(10)}{ - 8}$ ${x = 2}$ You can also plug ${y = 10}$ into ${7.5x+3y = 45}$ and get the same answer for $x$ ${7.5x + 3}{(10)}{= 45}$ ${x = 2}$ There were $2$ adults and $10$ kids.